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A263897
Number of anagram compositions of 2n that can be formed from the compositions of n.
2
1, 1, 2, 6, 16, 44, 134, 414, 1290, 4102, 13306, 43718, 145176, 487384, 1651154, 5636702, 19381392, 67074420, 233483430, 817106622, 2873589060, 10151255976, 36009769278, 128231072994, 458268615966, 1643227382022, 5910606009330, 21322486928518, 77132729929864
OFFSET
0,3
COMMENTS
A composition of n is an ordered sequence of positive integers whose sum is n. An anagram composition of n can be divided into two consecutive subsequences having the same parts, with a central part between the subsequences permitted.
a(n) is the number of anagram compositions of 2n with no central summand.
Here is a computation algorithm: List the individual partitions of n. Then for each partition, determine the corresponding number of compositions. Square each of these numbers then add them together.
LINKS
Miklos Bona, Rebecca Smith, An Involution on Involutions and a Generalization of Layered Permutations, arXiv preprint arXiv:1605.06158 [math.CO], 2016. See Theorem 4.3.
EXAMPLE
For n=3, the compositions are [3], [1,2], [2,1], [1,1,1]. The anagram compositions of 2*3=6 are [3][3], [1,2][1,2], [1,2][2,1], [2,1][1,2], [2,1][2,1], [1,1,1][1,1,1], so there are a(3)=6 anagram compositions in all.
For n=4, the individual partitions are [4], [3,1], [2,2], [2,1,1] and [1,1,1,1]. The corresponding number of compositions are 1, 2, 1, 3, and 1. The corresponding squares of these numbers are 1, 4, 1, 9, and 1. The sum of these squares is a(4)=16.
MAPLE
b:= proc(n, i, p) option remember; `if`(n=0, p!^2,
`if`(i<1, 0, add(b(n-i*j, i-1, p+j)/j!^2, j=0..n/i)))
end:
a:= n-> b(n$2, 0):
seq(a(n), n=0..35); # Alois P. Heinz, Oct 29 2015
MATHEMATICA
b[n_, i_, p_] := b[n, i, p] = If[n == 0, p!^2, If[i<1, 0, Sum[b[n-i*j, i-1, p+j]/j!^2, {j, 0, n/i}]]]; a[n_] := b[n, n, 0]; Table[a[n], {n, 0, 35}] (* Jean-François Alcover, Nov 05 2015, after Alois P. Heinz *)
PROG
(Python)
from sympy.core.cache import cacheit
from sympy import factorial as f
@cacheit
def b(n, i, p): return f(p)**2 if n==0 else 0 if i<1 else sum(b(n - i*j, i - 1, p + j)//f(j)**2 for j in range(n//i + 1))
def a(n): return b(n, n, 0)
print([a(n) for n in range(36)]) # Indranil Ghosh, Aug 18 2017, after Maple code
CROSSREFS
First differences of A097085.
Sequence in context: A105696 A074413 A373048 * A209629 A349488 A055544
KEYWORD
nonn
AUTHOR
Gregory L. Simay, Oct 28 2015
EXTENSIONS
a(9)-a(28) from Alois P. Heinz, Oct 29 2015
STATUS
approved